Step 1:
#- 0.8x + 0.6y = 2 # Let this be equation (1)
#3.2x - 0.4y = 4# Let this be equation (2)
Step 2:
To eliminate one variable , say #x# , we make the coefficient of that variable same in both equations by multiplying the equation by some constant.
here , we will multiply equation (1) by 4 to get
(1) x # 4 # gives ------- # -3.2 x + 2.4 y = 8#
Step 3: now the set of equation is
#-3.2 x + 2.4 y = 8# and
#3.2 x - 0.4 y = 4#
Step 4:
we observe that the coefficients of #x# in both equations are equal but of opposite sign. So we add these two equations in order to eliminate # x #
Adding the new set of equations we, have:
#(-3.2 x + 2.4 y) + (3.2 x - 0.4 y) = 8 + 4#
# -3.2 x + 2.4 y + 3.2 x - 0.4 y = 12# ---- open brackets
#-3.2 x + 3.2 x + 2.4 y - 0.4 y = 12# ---- grouping terms with same variables
# 0 x + 2.0 y =12#
# 2 y = 12#
#y = 6#
We get the value of #y =6#
Step 5:
Substitute this value of #y# in any one of the given equations to find value of #x#
Substituting #y =6# in equation (2):
#3.2x - 0.4y = 4#
# 3.2 x - 0.4 #x# 6 = 4 #
#3.2 x -2.4 = 4 #
#3.2 x = 4 + 2.4# ---- transposition
# 3.2 x = 6.4 #
# x = 2#
Therefore we have # x = 2 # and # y = 6#
Step 6: Cross check by substituting values of #x# and # y# in any one equation.
Substituting in equation (1), we have
#- 0.8x + 0.6y = 2 #
Left-hand-side of equation is
#- 0.8x + 0.6y #
#- 0.8# x # 2# + #0.6# x #6 #
# -1.6 + 3.6 #
# =2 # = Right- hand- side of equation.
Hence the obtained values are correct.
